Relative Frequency Histogram Percent Moeling Continuous 10 0 30 40 0 60 70 0 90 100 110 Test scores 1 Density function, f (x) Density Curve Percent A smooth curve that fit the istribution 10 0 30 40 0 60 70 0 90 100 110 Test scores Use a mathematical moel to escribe the variable. 3 Continuous Density Function (Curve) 1. Mathematical Formula Shows Probability Densities,, for All Values of x, & [ Is Not Probability ]. Property Total Area Uner Curve is 1. Probability Density Function (Value, Density) Value 4 x Continuous Ranom Variable P( c ) f ( x ) x c Meaning of Area Uner Curve : What percentage of the istribution is in between 7 an 6? Probability Is Area Uner Curve! c P(7) 0 7 6 (Height) 6-1
Uniform Skewe to right Symmetrical Skewe to left 1. Bell-Shape & Symmetrical. Mean, Meian, Moe Are Equal 3. Ranom Variable Has Infinite Range < x < f() Mean Meian Moe 7 Probability Density Curve Stanar 9 Probability Density Function 1 1 x f ( x ) e π f (x ) Density of Ranom Variable x Mean of the Stanar Deviation of the π 3.1419 ; e.71 x Value of Ranom Variable ( < x < ) Notation: N (, ) Α normal istribution with mean an stanar eviation 10 Effect of Varying Parameters ( & ) N (7, ) Α normal istribution with mean 7 an stanar eviation. Possible situations: Test scores, pulse rates, f() B N (130, 4) Α normal istribution with mean 130 an stanar eviation 4. Possible situations: Weight, Cholesterol levels, A C 11 1 -
Probability is area uner curve! Probability c P( c x ) f ( x) x Stanar Stanar : A normal istribution with mean 0 an stanar eviation 1. Notation: 1 c x 13 ~ N ( 0, 1) Cap letter 0 14 Area uner Stanar Curve P (1 < < 3) 0 z How to fin the proportion of the are uner the stanar normal curve below z or say P ( < z )? P ( > 3) 1 0 3 0 3 Use Stanar Table!!! 1 16 Stanar P( > 0.3) Area above.3.374 Areas in the upper tail of the stanar normal istribution.00.01.0 0.0.00.496.49 0.1.460.46. 0..1.417.413 0.3 0.3.3.37.374 17 1-3
Stanar Stanar P(0 < < 0.3) Area between 0 an.3.16 Areas in the upper tail of the stanar normal istribution.00.01.0 0.0.00.496.49 P(< 0.3) Area below.3.66 Areas in the upper tail of the stanar normal istribution.00.01.0 0.0.00.496.49 0.1.460.46. 0.1.460.46. 0..1.417.413 0.3.3.37.374 0.3 Area. -.374.16 19 0..1.417.413 0.3.3.37.374 0.3 Area 1 -.374.66 0 P ( -1.00 < < 1.00 ).6 P ( -.00 < <.00 ).341.341-1.00 0 1.00 -.00 0.00 1 P ( -3.00 < < 3.00 ) P ( -1.40 < <.33 ).909.419.490-3.00 0 3.00-1.40 0.33 3 4-4
Stanarize the 1 Stanarize the N (, ) a b N ( 0, 1) a 0 Stanar b 0 One table! a b P ( a < < b) P < < 6 10 Stanarizing For a normal istribution that has a mean an s.. 10, what percentage of the istribution is between an 6.? 10 Stanarizing 6..1 10 P( 6.) P(0.1) 1 6. 7 6. 0.1.00.01.0 0.0.00.496.49 0.1.460.46. 0..1.417.413 0.3.3.37.374 Obtaining the Probability Probability Table (Portion) 0 1..1 Area. -..04 9 10 3. P(3. ) 3..1 10 P(3. ) P(.1 0) 1.04 -.1 0 Area.04 30 -
10 P(.9 7.1).9.1 10 7.1.1 10 P(.9 7.1) P(.1.1) 1 10 P( > ).30 10 P( > ) P( >.30) 1.166.3.9 7.1 -.1 0.1 Area.03 +.03.166 31 0 Area.3.30 3 P( > ) More on 10 6% 3% Value is the 6 n percentile 33 The work hours per week for resients in Ohio has a normal istribution with hours & 9 hours. Fin the percentage of Ohio resients whose work hours are A. between & 60 hours. P( 60)? B. less than 0 hours. P( 0)? 34 00 9 P( 60)? 0 9 60 9 P( 60) P(0 ) 1.477 47.7%.477 00 9 P( 0)? 0.44 9 P( 0) P(.44) 0.007 0.7% 1 60 0 400.0 3 0.007 0 -.44 400 0.0 36-6
What is z given P( < z) ).0?.0 Fining Values.0 0 z.4 Upper Tail Area 1 -.0.0 z.4 Probability Table (Portion).04.0.64.61. 0.7.33.30.7 0..03.00.19 0.9.176.174.171 37 Fining Values : The weight of new born infants is normally istribute with a mean 7 lb an stanar eviation of 1. lb. Fin the 0th percentile. Area to the left of 0th percentile in 0.00. In the table there is a area value 0.00 corresponing to a z-score of.4. 0th percentile 7 +.4 x 1..00 lb 3 Fining Values : The Boy Mass Inex for a particular population is normally istribute with a mean an stanar eviation of 4. Fin the 0th percentile. Area to the left of 0th percentile in 0.00. In the table there is a area value 0.00 corresponing to a z-score of.4. Fining Values : The Boy Mass Inex for a particular population is normally istribute with a mean an stanar eviation of 4. Fin the 40th percentile..40 0.. 39 40 0th percentile +.4 x 4.36 Fining Values : The Boy Mass Inex for a particular population is normally istribute with a mean an stanar eviation of 4. Fin the 40th percentile. Area to the left of 40th percentile in 0.40. In the table there is a area value 0.40 corresponing to a z-score of.. Stanine Score (Stanar Nine) 1 3 4 6 7 9? % 4% 4% -1.7-1. -.7 -. 0..7 1. 1.7 41 40th percentile. x 4 1-7